Question

If $\frac{x - 2}{x + 5} > 2$, then $x\in$

• A. $(-12$, $5)$
• B. $(-12$, $-5)$
• C. $(-5$, $12)$
• D. $(5$, $12)$

Answer 1

Answer: (B)

We have $\frac{x - 2}{x + 5} > 2$
$\Rightarrow \quad \frac{x - 2}{x + 5} -2 > 0 \Rightarrow \quad \frac{-\left(x + 12\right)}{x + 5} > 0 \Rightarrow \quad \frac{x + 12}{x + 5} < 0$
$\Rightarrow \quad x+ 12 > 0$ and $x + 5 < 0$
or $\quad x + 12 < 0$ and $x + 5 > 0$
$\Rightarrow \quad x> -12$ and $x < -5$
or $\quad x < -12$ and $x > -5\quad$ (Not possible)
Therefore, $-12 < x < - 5$, i.e. $x \in$ $(-12$, $-5)$